112. 路径总和

112. 路径总和

难度简单 523
给你二叉树的根节点 root 和一个表示目标和的整数 targetSum ，判断该树中是否存在 根节点到叶子节点 的路径，这条路径上所有节点值相加等于目标和 targetSum 。
叶子节点 是指没有子节点的节点。

示例 1：

输入：root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
输出：true

广度优先

public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) {
            return false;
        }
        Queue<TreeNode> queNode = new LinkedList<TreeNode>();
        Queue<Integer> queVal = new LinkedList<Integer>();
        queNode.offer(root);
        queVal.offer(root.val);
        while (!queNode.isEmpty()) &#123;
            TreeNode now = queNode.poll();
            int temp = queVal.poll();
            if (now.left == null && now.right == null) &#123;
                if (temp == sum) &#123;
                    return true;
                &#125;
                continue;
            &#125;
            if (now.left != null) &#123;
                queNode.offer(now.left);
                queVal.offer(now.left.val + temp);
            &#125;
            if (now.right != null) &#123;
                queNode.offer(now.right);
                queVal.offer(now.right.val + temp);
            &#125;
        &#125;
        return false;
    &#125;

作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/path-sum/solution/lu-jing-zong-he-by-leetcode-solution/

递归

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) {
            return false;
        }
        if (root.left == null && root.right == null) {
            return sum == root.val;
        }
        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
    }

// 作者：LeetCode-Solution
// 链接：https://leetcode-cn.com/problems/path-sum/solution/lu-jing-zong-he-by-leetcode-solution/
}